Some Basic Concepts of Chemistry 1 Question 36
56. (a) $1.0 \mathrm{~L}$ of a mixture of $\mathrm{CO}$ and $\mathrm{CO}_{2}$ is taken. This mixture is passed through a tube containing red hot charcoal. The volume now becomes $1.6 \mathrm{~L}$. The volumes are measured under the same conditions. Find the composition of mixture by volume.
(b) A compound contains 28 per cent of nitrogen and 72 per cent of a metal by weight. 3 atoms of metal combine with 2 atoms of nitrogen. Find the atomic weight of metal.
$(1980,5 \mathrm{M})$
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Solution:
- (a) After passing through red-hot charcoal, following reaction occurs
$$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \longrightarrow 2 \mathrm{CO}(g) $$
If the $1.0 \mathrm{~L}$ original mixture contain $x$ litre of $\mathrm{CO}_{2}$, after passing from tube containing red-hot charcoal, the new volumes would be :
$2 x$ (volume of $\mathrm{CO}$ obtained from $\left.\mathrm{CO}_{2}\right)+1$
$$ \begin{aligned} & \ \Rightarrow \quad x \text { (original CO) } & =1+x=1.6 \text { (given) } \ x & =0.6 \end{aligned} $$
Hence, original $1.0 \mathrm{~L}$ mixture has $0.4 \mathrm{~L} \mathrm{CO}$ and $0.6 \mathrm{~L}^{-}$of $\mathrm{CO}{2}$, i.e. $40 % \mathrm{CO}$ and $60 % \mathrm{CO}{2}$ by volume.
(b) According to the given information, molecular formula of the compound is $M_{3} \mathrm{~N}_{2}$. Also, 1.0 mole of compound has $28 \mathrm{~g}$ of nitrogen. If $X$ is the molar mass of compound, then :
$$ \begin{gathered} X \times \frac{28}{100}=28 \ \Rightarrow \quad X=100=3 \times \text { Atomic weight of } M+28 \ \Rightarrow \quad \text { Atomic weight of } M=\frac{72}{3}=24 \end{gathered} $$
